Quadrilaterals MA 341 Topics in Geometry Lecture 23
Theorems 1. A convex quadrilateral is cyclic if and only if opposite angles are supplementary. (Circumcircle, maltitudes, anticenter) 2. A convex quadrilateral is tangential if and only if opposite sides sum to the same measure. (Incircle, incenter) 24-Oct-2011 MA 341 001 2
Bicentric Quadrilaterals A convex quadrilateral is bicentric if it is both cyclic and tangential. A bicentric quadrilateral has both a circumcircle and an incircle. A convex quadrilateral is bicentric if and only if a + c = b + d and A + C = 180 = B + D 24-Oct-2011 MA 341 001 3
Bicentric Quadrilaterals 24-Oct-2011 MA 341 001 4
Bicentric Quadrilaterals X A, X B, X C, X D points of contact of incircle and quadrilateral Bicentric iff X A X C X B X D or AX A DXC or = XB A XC C AC AX = A+ CXC BC BX +DX B D 24-Oct-2011 MA 341 001 5
Bicentric Quadrilaterals M 1, M 2, M 3, M 4 midpoints of X A X B, X B X C, X C X D, X D X A Bicentric iff M 1 M 2 M 3 M 4 is a rectangle. 24-Oct-2011 MA 341 001 6
Newton Line Theorem: (Léon Anne) Let ABCD be a quadrilateral that is not a parallelogram. P lies on the line joining the midpoints of the diagonals iff K K =K K APB CPD BPC APD 24-Oct-2011 MA 341 001 7
Proof Drop perpendicular from B & D to EF. 24-Oct-2011 MA 341 001 8
Proof E midpoint of BD, EPB= DEX (vertical angles), ΔEPB= ΔDEX (Hyp-ang) perpendiculars are equal K DPF = K BPF. 24-Oct-2011 MA 341 001 9
Likewise K APF = K CPF. Proof 24-Oct-2011 MA 341 001 10
Proof K APB = K AFB + K APF + K BPF 24-Oct-2011 MA 341 001 11
Proof K DPC = K DFC -K CPF -K DPF 24-Oct-2011 MA 341 001 12
Proof K APB = K AFB + K APF + K BPF K DPC = K DFC -K CPF K DPF So K APB + K DPC =K AFB + K DFC 24-Oct-2011 MA 341 001 13
Proof K APD = K AFD + K DPF + K APF K BPC = K BFC -K BPF K CPF So K APD + K BPC =K AFD + K CFB 24-Oct-2011 MA 341 001 14
Proof K APB + K DPC = K AFB + K DFC K APD + K BPC = K AFD + K CFB Also K AFB = K CFB (equal bases, same height) and K DFC = K AFD (equal bases, same height) So K APB + K DPC = K APD + K BPC 24-Oct-2011 MA 341 001 15
Newton Line Theorem: The center of the circle inscribed into a quadrilateral lies on the line joining the midpoints of the diagonals. 24-Oct-2011 MA 341 001 16
Proof The distance from O to each side is R. The area of each triangle then is K AOB = ½ R AB K BOC = ½ R BC K COD = ½ R CD K DOA = ½ R AD Since AB + CD = BC + AD, multiplying both sides by ½R, we get that K AOB + K COD = K BOC + K DOA Thus, O lies on the Newton Line. 24-Oct-2011 MA 341 001 17
Area A bicentric quadrilateral is a cyclic quadrilateral, so Brahmagupta s Formula applies: K = (s - a)(s -b)(s -c)(s -d) Since it is a tangential quadrilateral we know that a + c = s = b + d. Thus: s a = c, s b = d, s c = a and s d = b or K= abcd 24-Oct-2011 MA 341 001 18
k,l = tangency chords p,q = diagonals klpq K= k +l 2 2 Area 24-Oct-2011 MA 341 001 19
Area r = inradius R = circumradius 4r K 2R 2 2 24-Oct-2011 MA 341 001 20
Inradius & Circumradius Since it is a tangential quadrilateral r K a+c Since it is bicentric, we get r abcd a+c We gain little for the circumradius 1 (ab + cd)(ac +bd)(ad +bc) R 4 abcd 24-Oct-2011 MA 341 001 21
Fuss Theorem Let x denote the distance from the incenter, I, and circumcenter, O. Then 1 1 1 + = (R - x) (R + x) r 2 2 2 24-Oct-2011 MA 341 001 22
NOTE Let d denote the distance from the incenter, I, and circumcenter, O, in a triangle, then Euler s formula says or 1 1 1 + = R-d R+d r 2 2 2Rr =R -d 24-Oct-2011 MA 341 001 23
Proof Let K,L be points of tangency of incircle A + C = 180 IL = r = IK and ILC = IKA = 90 Join ΔILC and ΔIKA to form ΔCIA I A K,L C 24-Oct-2011 MA 341 001 24
Proof What do we know about ΔCIA? A + C = 180 LCI = ½C and IAK= ½A LCI + IAK = 90 CIA = 90 ΔCIA a right triangle Hypotenuse = AK+CL I A K,L C 24-Oct-2011 MA 341 001 25
Proof Area = ½r(AK+CL) = ½ AI CI or r(ak+cl) = AI CI Pythagorean Theorem gives (AK+CL) 2 = AI 2 + CI 2 From first equation we have r 2 (AI 2 + CI 2 ) = AI 2 CI 2 1 1 1 = + r AI CI 2 2 2 A I K,L C 24-Oct-2011 MA 341 001 26
Proof Extend AI and CI to meet circumcircle at F and E. DOF+ DOE = 2 DAF + 2 DCE = BAD + BCD = 180 EF is a diameter! 24-Oct-2011 MA 341 001 27
Proof X = IO = median in ΔIFE. 2 1 2 2 1 2 IO = (IE + IF ) - EF 2 4 1 1 IO = (IE + IF ) - (2R) 2 4 2 2 2 2 IE + IF 2IO 2R 2 2 2 2 2 2 =2(x R ) 24-Oct-2011 MA 341 001 28
Proof From chords CE and AF, we have CI FI AI = EI AI FI = CI EI From chords CE and XY CI EI = XI YI =(R + x)(r - x) 2 2 =R -x 24-Oct-2011 MA 341 001 29
AI FI =R - x 2 2 Proof 2 2 1 1 FI EI + = + AI CI CI EI CI EI 2 2 2 2 2 2 EI +FI = (R - x ) 2 2 2 2 2 2 2 2(R + x ) = (R - x ) 2 2 2 AI FI = CI EI 24-Oct-2011 MA 341 001 30
Proof 2 2 1 2(R +x ) = 2 2 2 2 r (R -x ) ((R + x ) 2 +(R - x) 2 ) = 2 2 2 (R - x ) 1 1 1 = + 2 2 2 r (R+x) (R-x) 2 2 2 2 2 2 2r (R + x ) =(R - x ) 2 2 2 2 x= R +r -r 4R +r 24-Oct-2011 MA 341 001 31
Other results In a bicentric quadrilateral the circumcenter, the incenter and the intersection of the diagonals are collinear. Given two concentric circles with radii R and r and distance x between their centers satisfying the condition in Fuss' theorem, there exists a convex quadrilateral inscribed in one of them and tangent to the other. 24-Oct-2011 MA 341 001 32
Other results Poncelet s Closure Theorem: If two circles, one within the other, are the incircle and the circumcircle of a bicentric quadrilateral, then every point on the circumcircle is the vertex of a bicentric quadrilateral having the same incircle and circumcircle. 24-Oct-2011 MA 341 001 33